hw4.sol.Spring07

# hw4.sol.Spring07 - ME 452: Machine Design II Solution to...

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ME 452: Machine Design II Solution to Homework Set 4. Problem 1. (Problem 6-21, page 342) The minimum ultimate tensile strength can be found using Equation (2-17), page 37, as follows: 3.41 3.41 490 1671 MPa ut B SH == × = The problem statement suggests that the yield strength is 90 percent of the ultimate strength. Therefore 0.9 0.9 1671 1504 MPa yu t SS × = Because 1400 MPa ut S > , then from Equation (6-8), page 274 we get: 700 MPa e S = From Table (6-2), page 280, for ground surface finish we have: 1.58 a = and 0.085 b = − Therefore from Equation (6-19), page 279, the surface condition modification factor is given by: 0.085 1.58 (1671) 0.841 b au t ka S ==× = For the given cross-section, the equivalent diameter is given by Equation (6-25), page 281 as: 1/2 0.808( ) e dh b = Therefore 0.808(3 18) 5.938 mm e d = Then the size factor is given by Equation (6-20), page 280 as: 0.107 0.107 1.24 1.24 (5.938) 1.025 b kd −− × =

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The loading factor from Equation (6-26), page 282 is given by (also see page 306 for discussion on combined loading): 1 c k = Assuming that the operating temperature is 70 F ° the temperature factor from Table (6- 4), page 283, is given by: 1 d k = Assuming 50% reliability, the reliability factor from Table (6-5), page 285 is given by: 1 e k = Assuming there are no miscellaneous effects (see Page3 285 and 286), the miscellaneous- effects factor is given by: 1 f k = The endurance strength is then given by the Marin equation (see Equation 6-18, page 279) : ea b c d e f e Sk k k k k k S = Then 0.841 1.025 1 1 1 1 700 603.4 MPa e S = × ××××× = (a). The moment of inertia for the spring is given by () 34 1 18 3 40.5 mm 12 I == The deflection of the spring at the point of application of the force F is given by 3 3 Fl y EI = Therefore 3 3 EIy F l = where 100 mm l = is the length of the spring. From Table (A-5), page 987 the modulus of elasticity for carbon steel is given by 207 GPa E =
The minimum latching force is given by the initial deflection of min 2 mm y = 91 2 3 min min 33 9 3 3 207 10 40.5 10 2 10 50.3 N 100 10 EIy F l −− ×× × × × × == = × The maximum latching force is given by the maximum deflection of max 246 m m y =+= 2 3 max max 9 3 3 207 10 40.5 10 6 10 150.9 N 100 10 EIy F l × × × × = × (b).

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## This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

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hw4.sol.Spring07 - ME 452: Machine Design II Solution to...

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